Why am I losing solutions when solving Inequalities like $\frac{2y-3}{y}>0$ by multiplying both sides by $y$? So, I need to solve the following Inequality-
$$\frac{2y-3}{y}>0$$
I proceeded in the following manner-
$$2y-3>0\qquad   \text{[Multiplying both sides by }y.]$$
$$y>\frac{3}{2}\qquad  \text{[Adding 3 to both sides and and dividing by 2.]}$$
So my solution set is ($\frac{3}{2},\infty$)
But clearly, the inequality also holds true whenever y has any negative value. Hence, my solution set should in fact be: $(-\infty,0)\cup(\frac{3}{2},\infty)$.
My question is- What is the proper way of solving such inequalities in order to get all values of the variable? I am looking for a simple explanation(pre-college level) if possible.
 A: 
[Multiplying both sides by y.]

Here. You multiplied both sides of an inequality by some number, but you forgot to check whether that number was positive (in which case the sign remains ">") or negative (in which case the sign becomes "<").
If you want to make sure you have all the solutions, make sure you don't accidentally multiply or divide by something that might be non-positive, or each time you do it, split it into two cases ("Case 1: y > 0" and "Case 2: y < 0").
A: Consider the following general inequality:
$$\frac{f(y)}{g(y)} > 0.$$
Suppose that:


*

*$f(y) > 0$ for all $y \in F \subseteq \mathbb{R}$;

*$g(y) > 0$ for all $y \in G \subseteq \mathbb{R}$.


Starting from the sets $F$ and $G$, you must build up a third set as follows:
$$H = \{y \in \mathbb{R} : [f(y)>0 ~\text{and}~ g(y)>0]~\text{or}~[f(y)<0 ~\text{and}~ g(y)<0]\}.$$
This means that $y \in H$ if one of the following holds true:


*

*$y \in F$ and $y \in G$;

*$y \not\in F$ and $y \not\in G$.


For the particular case:
$$\frac{2y-3}{y}> 0.$$
$$f(y) = 2y-3>0 \Rightarrow F = \left\{y > \frac{3}{2}\right\}.$$
$$g(y) = y>0 \Rightarrow G = \left\{y > 0\right\}.$$
Then:


*

*If $y>\frac{3}{2}$, then $y \in F$ and $y \in G$. Therefore, $y \in H$.

*If $y < 0$, then $y \not\in F$ and $y \not\in G$. Therefore, $y \in H$.


Finally, the set H is:
$$H = \left\{y > \frac{3}{2} \vee y < 0\right\}.$$
A: If $y<0$ (clearly valid), when you multiply both sides by $y$, the inequality flips over; however, you stick to only one case of possible results. 
One trick to get rid of this, is to multiply both sides by $y^2$, which is guaranteed to be positive: $$y(2y-3)>0\implies y<0\quad\text{or}\quad y>3/2$$
A: The correct way to do it is.
Case 1: $y > 0$. then $2y - 3 > 0$ and $y > 0$.
Case 2: $y = 0$.  This is impossible.
Case 3: $y < 0$ and $2y - 3 < 0$ and $y < 0$.
In case 1: $2y -3 > 0$ so 
$2y > 3$
$y > \frac 32$ and $y > 0$ so $y> \frac 32 > 0$
OR
Case 3: $2y - 3< 0$ so
$2y < 3$
$y < \frac 32$ and $y < 0$.  So $y < 0 < \frac 32$.
So either $y > \frac 32$ OR $y < 0$.
